Register
1. The heights of American adult males are normally distributed with a mean of 177 cm and a standard deviation of 7.4 cm. Use the Empirical Rule to find the range of heights that contain approximately(a)
133k views
2 votes
1. The heights of American adult males are normally distributed with a mean of 177 cm and a standard deviation of 7.4 cm. Use the Empirical Rule to find the range of heights that contain approximately(a) 68% of the data cm -- cm(b) 95% of the data cm -- cm(c) 99.7% of the data cm -- cm

1. The heights of American adult males are normally distributed with a mean of 177 cm-example-1
User Evildead
by
2.8k points

2 Answers

7 votes

If the heights of American adult males are normally distributed with a mean of 177 cm and a standard deviation of 7.4 cm. Tthe range of heights that contain:

(a) 68% of the data falls within the range of 169.6 cm - 184.4 cm.

(b) 95% of the data falls within the range of 162.2 cm - 191.8 cm.

(c) 99.7% of the data falls within the range of 154.8 cm - 199.2 cm.

What is Range of heights?

(a) Range of heights:

Lower range: Mean - 1 standard deviation

Lower range = 177 - 7.4 = 169.6 cm

Upper range: Mean + 1 standard deviation

Upper range = 177 + 7.4 = 184.4 cm

(b) Range of heights:

Lower range: Mean - 2 standard deviations

Lower range = 177 - (2 * 7.4) = 162.2 cm

Upper range: Mean + 2 standard deviations

Upper range = 177 + (2 * 7.4) = 191.8 cm

(c) Range of heights:

Lower range: Mean - 3 standard deviations

Lower range = 177 - (3 * 7.4) = 154.8 cm

Upper range: Mean + 3 standard deviations

Upper range = 177 + (3 * 7.4) = 199.2 cm

Therefore:

(a) 68% of the data falls within the range of 169.6 cm - 184.4 cm.

(b) 95% of the data falls within the range of 162.2 cm - 191.8 cm.

(c) 99.7% of the data falls within the range of 154.8 cm - 199.2 cm.

User Benjamin Li
by
3.3k points
2 votes

\begin{gathered} \operatorname{mean}\text{ = }\mu=177 \\ s\tan dart\text{ }deviation\text{ = }\sigma=7.4 \\ \text{Emperical rule for }68\text{\%} \\ heights=\mu\pm1\cdot\sigma \\ heights=177\pm1\cdot(7.4) \\ \text{height1 = 177+7.4=184.4} \\ height\text{ 2 =177-7.4=169.6} \\ 68\text{\% of the data } \\ 169.6\operatorname{cm}--184.4\operatorname{cm} \\ \text{Emperical rule for }95\text{\%} \\ heights=\mu\pm2\cdot\sigma \\ \text{heights}=177\pm2\cdot(7.4) \\ \text{height1 = 177+14.8=1}91.8 \\ \text{height2 = 177-14.8=162.2} \\ 95\text{\% of the data } \\ 162.2cm--191.8\operatorname{cm} \\ \text{Emperical rule for }99.7\text{\%} \\ heights=\mu\pm3\cdot\sigma \\ \text{heights}=177\pm3\cdot(7.4) \\ \text{height1 =177+22.2=1}99.2 \\ \text{height2 =}177-22.2=154.8 \\ 99.7\text{\% of the data } \\ 154.8\operatorname{cm}--199.2\operatorname{cm} \end{gathered}

User Finn Larsen
by
3.8k points
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

3.3m questions

4.2m answers

Other Questions

A study was completed in Florida. In southern Florida, the study involved 3,000 patients; 54% of them experienced flulike symptoms during the same month. The study had a margin of error of 1.8%. What does
The sanchez family goes out for dinner, and the price of the meals is $60. The sales tax on the Meals is 7 percent, and they also want to leave a 15 percent tip. What is the total cost of the meal? Analyze
Simplify: 10 ÷ [(5 + 2) − 5]
WHAT IS 33 TIMES 5 TELL ME THE ANWER
At the beginning of a population study, a city had 240,000 people. Each year since, the population has grown by 9.7%. Lett be the number of years since start of the study. Let y be the city's population.
Link Copied!